I did a lab and below was the procedure. I need to find a molar concentration of

Question

I did a lab and below was the procedure. I need to find a molar concentration of the Na2S2O3 solution.

Preparation of standard 0.02 M KIO3 solution - Accurately weigh by difference about 2.0 g of the solid KIO3 into a small beaker and transfer it quantitatively to a 500 mL volumetric flask. Add water and stir to dissolve solid. Dilute to the mark and mix thoroughly. Using the mass of KIO3 you weighed-out calculate and record the molar concentration of the solution.

--> The molar concentration I got was 0.0202M by adding 2.1595g of solid KIO3.

Preparation and standardization of 0.1 M Na2S2O3 - Boil about 500 mL of distilled water for 10 to 15 min. Allow the water to cool to room temperature; then add about 12 g of Na2S2O3·5H2O and 0.05 g of Na2CO3. Stir until the solid has dissolved. Transfer the solution to a clean glass or plastic bottle and store in a dark place.

--> I added 12.0373g of Na2S2O3·5H2O and 0.0497g of Na2CO3.

Pipet 25 mL of the KIO3 solution into each of three 250 mL Erlenmeyer flasks. From this point treat each of the flasks separately until the titration is finished. Add 2 g of solid KI and 10 mL of 0.5 M H2SO4 to the first flask and immediately begin titrating with the Na2S2O3 solution.

--> I added 2.0074g of KI

When the titration solution has lost almost all of its color (only a pale yellow remains), add 2 mL of starch indicator. Continue titrating until the blue color disappears.

--> The titration volume was 0.0311L

Calculate and record the molar concentration of the solution.

Explanation / Answer

For KIO3: Concentration = M1 = 0.0202 M

Volume = V1 = 25 ml

For Na2S2O3, concentration = M2 = ?

Volume = V2 = 0.0311 L = 31.1 ml

In reaction of KIO3 ans Na2S2O3, 1 moles of Na2S2O3 reacts with 2 mol of KIO3

So, M1V1 = 2(M2V2)

0.0202*25 = 2(M2*31.1)

M2 = 0.0162 M

Concentration of Na2S2O3 = 0.0162 M

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